$(X,\Sigma,\mu)$ be a probability space, $A\subset\Sigma$ be a field s.t. $\Sigma=\sigma(A)$.Prove inf$\{\mu(E\Delta F):F\in A\}=0\forall E\in\Sigma$ 
$(X,\Sigma,\mu)$ be a probability space, $\mathcal{A}\subset\Sigma$ be a field
such that $\Sigma=\sigma(A)$. Prove that, for all $E\in\Sigma$ $$\text{inf }\{\mu(E\Delta F):\ F\in \mathcal{A}\}=0$$

I tried this using Monotone Class Theorem. Define $\mathcal{M}=\{E\in\Sigma|\ \text{inf }\{\mu(E\Delta F):\ F\in\mathcal{A}\}=0 \}$. It's easy to see that $\mathcal{M}$ contains $\mathcal{A}$. As $\sigma(\mathcal{A})=\Sigma$, by Monotone class theorem it's enough to prove $\mathcal{M}$ is monotone class.
So let $E_n\in\mathcal{M}$ such that $E_n\nearrow E$. We have to show that $E\in\mathcal{M}$. We have $\text{inf }\{\mu(E_n\Delta F):\ F\in\mathcal{A}\}=0\ \forall n$. Choose $\epsilon>0$. To find $F\in\mathcal{A}$ such that $\mu(E\Delta F)<\epsilon$. We know that as $\mu$ is a finite measure, for FIXED $F\in\mathcal{A}$, $\mu(E_n\Delta F)\to 0$.
Now $\text{inf }\{\mu(E_n\Delta F):\ F\in\mathcal{A}\}=0\implies \exists F\in\mathcal{A}$ such that $\mu(E_n\Delta F)<\epsilon$. But here the problem is $F$ depends upon $n$, so I'm unable to apply $\mu(E_n\Delta F)\to 0$.
Can anyone give me any idea how to argue it? Thanks for help in advance.
 A: Hints: Choose $F_n \in \mathcal A$ such that $\mu (E_n\Delta F_n)<\frac {\epsilon} {2^{n}}$. Check that $(\bigcup E_n) \Delta (\bigcup F_n )\subseteq \bigcup (E_n \Delta F_n)$. Of course, $\bigcup F_n$ need not be in $\mathcal A$. So we have to replace it with $\bigcup\limits_1^{N} F_n$ with large enough $N$.  Since $\mu$ is a  finite measure $\mu(\bigcup\limits_{1}^{N} F_n) \to \mu(\bigcup F_n )$ as $N \to \infty$.
A: I think one can prove directly that $\mathcal{M}$ is a $\sigma$-algebra.

*

*Given $\varepsilon>0$, If $A\in\mathcal{M}$ then there is $A'\in\mathcal{A}$ such that $\mu(A\triangle A')<\varepsilon$. Since $A\triangle A'=A^c\triangle (A')^c$, and $(A')^c\in\mathcal{A}$, it follows that $A^c\in\mathcal{A}$.

*Suppose $(A_n:n\in\mathbb{N})\subset\mathcal{M}$. Given $\varepsilon$, there is for each $n\in\mathbb{N}$ a set $A'_n\in\mathcal{A}$ such that
$\mu(A_n\triangle A'_n)<2^{-n-1}\varepsilon$.
Now, as $\lim_N\mu\Big(\bigcup^N_{k=1}A_k\Big)=\mu\Big(\bigcup_nA_n\Big)$, there is $N$ large enough such that
$$\mu\Big(\big(\bigcup_nA_n\big)\setminus\big(\bigcup^N_{k=1}A_k\big)\Big)<\varepsilon/2$$
Then $$\begin{align}
\mu\Big(\big(\bigcup_nA_n\big)\triangle\big(\bigcup^N_{k=1}A'_k\big)\Big)& \leq \mu\Big(\big(\bigcup_nA_n\big)\setminus\big(\bigcup^N_{k=1}A_k\big)\Big)+\mu\Big(\big(\bigcup^N_{j=1}A_j\big)\triangle\big(\bigcup^N_{k=1}A'_k\big)\Big)\\
&<\varepsilon/2 +\mu\Big(\bigcup^N_{j=1}A_j\triangle A'_j\big)\leq\frac{\varepsilon}{2}+\sum^N_{j=1}\frac{\varepsilon}{2^{n+1}}<\varepsilon
\end{align}$$
This shows that $\bigcup_nA_n\in\mathcal{M}$

*Clearly $\mathcal{A}\subset \mathcal{M}$ and so, $X\in\mathcal{M}$.

Consequently $\mathcal{M}$ is a field and as $\mathcal{A}\subset\mathcal{M}\subset=\sigma(\mathcal{A})=\Sigma$, we conclude that $\mathcal{M}=\Sigma$.

In this posting I have used the following obvious facts:

*

*$\mathbb{1}_{A\triangle B} =|\mathbb{1}_A-\mathbb{1}_B|$
From this is clear that

*If $A\subset B$, then $B\setminus A=B\triangle A$

*For any sets $A,B, C$, $\mathbb{1}_{A\triangle B}\leq \mathbb{1}_{A\triangle C} + \mathbb{1}_{C\triangle B}$. This in particular implies that  $\mu(A\triangle B)\leq \mu(A\triangle C) +\mu(C\triangle B)$ for all $A,B,C\in\Sigma$.

Finally,
4. For any set  $A_1,\ldots, A_n$ and $B_1,\ldots B_n$
$$\begin{align}\big(\bigcup^n_{j=1}A_n \big)\triangle\big(\bigcup^n_{k=1}B_k\big)&=
\Big(\big(\bigcup^n_{j=1}A_j\big)\cap\big(\bigcup^n_{k=1}B_k\big)^c\Big)\cup\Big(\big(\bigcup^n_{k=1}B_k\big)\cap\big(\bigcup^n_{j=1}A_j\big)^c\Big)\\
&=\Big(\bigcup^n_{j=1}\big(A_j\cap\bigcap^n_{k=1}B^c_k\big)\Big)\cup\Big(\bigcup^n_{k=1}\big(B_k\cap\bigcap^n_{j=1}A^c_j\big)\Big)\\
&\subset\Big(\bigcup^n_{j=1}(A_j\setminus B_j)\Big)\cup\Big(\bigcup^n_{k=1}B_k\setminus A_k\Big)\\
&=\bigcup^n_{j=1}A_j\triangle B_j
\end{align}$$
This implies that
$$\mu\Big(\big(\bigcup^n_{j=1}A_n\big)\triangle\big(\bigcup^n_{k=1}B_n\big)\Big)\leq \sum^n_{j=1}\mu(A_n\triangle B_n)$$
